Question

A thin aluminum sheet with an emissivity of \(0.15\) on both sides is placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=900 \mathrm{~K}\) and \(T_{2}=650 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.8\), respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result with that without the shield.

Short answer

Answer: The heat transfer ratio between the two plates with the shield in place compared to without the shield is approximately 0.263, or 26.3%.

Step-by-step solution

Find the effective emissivities between the surfaces.

We need to use the equation for effective emissivity between surfaces, given by: \(\frac{1}{\varepsilon_{\text{effective,1-2}}} = \frac{1-\varepsilon_1}{\varepsilon_1 \mathcal{A}_1} +\frac{\mathcal{A}_1}{\mathcal{A}_2}\frac{1-\varepsilon_2}{\varepsilon_2}\) In this case, the area of both plates is the same, so \(\mathcal{A}_1=\mathcal{A}_2\), and the equation simplifies to: \(\varepsilon_{\text{effective,1-2}} = \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2-\varepsilon_1 \varepsilon_2}\) Now we can plug in the given emissivity values to find the effective emissivity between the plates without the shield: \(\varepsilon_{\text{effective,1-2}} = \frac{(0.5)(0.8)}{0.5+0.8-(0.5)(0.8)}\) We get: \(\varepsilon_{\text{effective,1-2}} = 0.615\) With the shield in place, we have to modify our equation to account for the additional surface: \(\frac{1}{\varepsilon_{\text{effective,1-3}}} = \frac{1-\varepsilon_1}{\varepsilon_1 \mathcal{A}_1} +\frac{\mathcal{A}_1}{\mathcal{A}_{s}}\frac{1-\varepsilon_s}{\varepsilon_s}+\frac{\mathcal{A}_{s}}{\mathcal{A}_2}\frac{1-\varepsilon_2}{\varepsilon_2}\) For this case, the areas of the plates are also the same as the shield's (\(\mathcal{A}_1=\mathcal{A}_{s}=\mathcal{A}_2\)), and we can plug in the given emissivity values: \(\varepsilon_{\text{effective,1-3}} = \frac{(0.5)(0.15)(0.8)}{[(0.5)(0.15)+0.15+0.15^2+(0.15)(0.8)]}\) We get: \(\varepsilon_{\text{effective,1-3}} = 0.162\)

Calculate the radiation heat transfer between the plates.

Now we can use the formula for radiation heat transfer per unit surface area of the plates: \(q = \varepsilon_{\text{effective}}\sigma (T_1^4-T_2^4)\) Plug in the given temperatures and the calculated effective emissivities to find the heat transfer for both cases: \(q_{\text{Without Shield}} = 0.615\sigma (900^4-650^4)\) \(q_{\text{With Shield}} = 0.162\sigma (900^4-650^4)\)

Compare the results for both cases.

Divide the heat transfer rate with and without the shield to find the ratio: \(\text{Heat Transfer Ratio} = \frac{q_{\text{With Shield}}}{q_{\text{Without Shield}}} = \frac{0.162}{0.615}\) We get the heat transfer ratio as: \(\text{Heat Transfer Ratio} \approx 0.263\) So, by comparing the results, we find that with the shield in place, the rate of radiation heat transfer between the two plates is only about 26.3% of the heat transfer without the shield.

Key concepts

Effective Emissivity

In radiation heat transfer, effective emissivity is a crucial concept for understanding how well heat is transferred between surfaces. It represents the combined effect of multiple surfaces and their emissivities on the thermal exchange process. When dealing with two large parallel plates, effective emissivity is calculated to understand how they interact thermally. If the plates have the same area, the effective emissivity can be calculated using the formula: \[ \varepsilon_{\text{effective,1-2}} = \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1 + \varepsilon_2 - \varepsilon_1 \varepsilon_2} \] This equation helps assess how much heat is exchanged between the surfaces, factoring in each material's properties. By plugging in the emissivity values for the specific surfaces, we get a comprehensive view of their thermal interactions. Understanding this concept helps predict how surfaces will heat or cool, impacting various engineering applications.

Radiation Shielding

Radiation shielding is an effective method to reduce heat transfer, especially between two bodies with high-temperature differences. Placing a shield between two surfaces alters the thermal interaction by creating an additional step for the heat transfer, thus slowing down the process. In the exercise given, using a thin aluminum sheet with known emissivity between two plates demonstrates how an extra barrier can decrease the effective emissivity and, consequently, the heat transfer between the surfaces. The shield's effectiveness derives from its ability to reflect and absorb some of the radiation, reducing the net transfer of heat. By comparing scenarios with and without a shield, we observe that the heat transfer rate significantly decreases, showcasing the shield's practicality in thermal management applications.

Emissivity Calculation

Calculating emissivity is vital for determining how well a surface can emit thermal radiation. In the context of our exercise, various calculations illustrate how different emissivities affect overall heat exchange. For example, the thin aluminum sheet has a specific emissivity, which, in combination with the plates' emissivities, determines the effective emissivity of the entire setup. Utilizing correct formulas and plugging in respective emissivity values mapped out in the exercise allows for precision in the thermal analysis. This calculation is critical for engineers to evaluate the energy conservation and efficiency in systems where thermal radiation is predominant. By calculating the effective emissivity accurately, we get insights into how modifications, such as adding shields, affect thermal performance.

Heat Transfer Reduction

Minimizing heat transfer is often a necessity in thermal management systems. In situations where controlling temperature is critical, understanding how to reduce radiation heat transfer can lead to substantial energy savings. Placing a radiation shield, such as the aluminum sheet in the exercise, between two plates is a practical strategy for reducing heat transfers. The effectiveness is quantified by calculating the heat transfer rates with and without the shield, typically showing a significant reduction when a shield is used. This concept is instrumental in designing thermal control systems, ensuring that heat does not move unchecked where it can cause disruptions or energy inefficiencies.

Thermal Radiation Analysis

Analyzing thermal radiation involves understanding how heat is transferred via electromagnetic radiation between surfaces at different temperatures. This involves careful calculation of effective emissivities and appreciating the impact of shields on the thermal interactions between surfaces, as demonstrated in our scenario. The detailed analysis helps not only in understanding the direct effects but also in foreseeing the thermal behavior of entire systems under varying conditions. Moreover, by comparing different setups—such as those with an added radiation shield—engineers and scientists can deduce practical solutions to real-world thermal management challenges. By dissecting these complex processes into simpler calculations, we gain comprehensive insights into optimizing energy efficiency in both industrial and daily applications.